Problem: Solve for $q$, $ \dfrac{9}{4q^3} = \dfrac{2q + 3}{q^3} - \dfrac{7}{q^3} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4q^3$ $q^3$ and $q^3$ The common denominator is $4q^3$ The denominator of the first term is already $4q^3$ , so we don't need to change it. To get $4q^3$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{2q + 3}{q^3} \times \dfrac{4}{4} = \dfrac{8q + 12}{4q^3} $ To get $4q^3$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{7}{q^3} \times \dfrac{4}{4} = -\dfrac{28}{4q^3} $ This give us: $ \dfrac{9}{4q^3} = \dfrac{8q + 12}{4q^3} - \dfrac{28}{4q^3} $ If we multiply both sides of the equation by $4q^3$ , we get: $ 9 = 8q + 12 - 28$ $ 9 = 8q - 16$ $ 25 = 8q $ $ q = \dfrac{25}{8}$